3
$\begingroup$

How would I find the summation of

$$\sum\limits_{x = 3}^{\infty} 1.536 (x^2) \left(\frac 5 8\right)^x $$

Would I have to take the 2nd derivative of $(1/1-x)$?

$\endgroup$
5
$\begingroup$

Yes, but it’s helpful to insert an intermediate step. Starting with

$$\frac1{1-x}=f(x)=\sum_{n\ge 0}x^n\;,$$

you get

$$\frac1{(1-x)^2}=f\,'(x)=\sum_{n\ge 0}nx^{n-1}\;,$$

and therefore

$$\frac{x}{(1-x)^2}=xf\,'(x)=x\sum_{n\ge 0}nx^{n-1}=\sum_{n\ge 0}nx^n\;.$$

Now what happens if you differentiate a second time and then multiply by $x$ again?

$\endgroup$
  • $\begingroup$ Would I get -2x^2/(1-x)^3? $\endgroup$ – Jesus Oct 14 '13 at 2:21
  • $\begingroup$ @Jesus: Your differentiation is a bit off. I get $$\frac{1-x}{(1-x)^3}$$ for the derivative of $$\frac{x}{(1-x)^2}\;.$$ $\endgroup$ – Brian M. Scott Oct 14 '13 at 2:24
  • $\begingroup$ Wouldn't the answer be negative? Otherwise, x is the ratio (5/8) and I would need to subtract the first 2 terms from it which would be (1.536 * 5/8) and (1.536 * 4 * (5/8)^2)? $\endgroup$ – Jesus Oct 14 '13 at 2:30
  • $\begingroup$ @Jesus: No, the derivative is not negative. Yes, you’ll substitute $\frac58$ for $x$, and you’ll have to subtract the first three terms, the ones for $n=0$, $n=1$, and $n=2$. $\endgroup$ – Brian M. Scott Oct 14 '13 at 2:33
  • $\begingroup$ WolframAlpha gives me a negative derivative wolframalpha.com/input/?i=derive+x%2F%281-x%29%5E2... when n=0, wouldn't just be 0? Are the other 2 terms correct for when n=1 and n=2? $\endgroup$ – Jesus Oct 14 '13 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.